Surveying If θ and δ be the latitude of an observer and declination of a heavenly body respectively, the upper culmination of the body will be south of zenith if its zenith distance, is ½ (θ - δ) θ - δ δ - θ θ + δ ½ (θ - δ) θ - δ δ - θ θ + δ ANSWER DOWNLOAD EXAMIANS APP
Surveying The latitude (λ) of a place and the altitude (α) of the pole are related by λ = α - 90° λ = 180° - α λ = 90° - α λ = α λ = α - 90° λ = 180° - α λ = 90° - α λ = α ANSWER DOWNLOAD EXAMIANS APP
Surveying The altitude of a heavenly body is its angular distance, measured on the vertical circle passing through the body, above Pole Equator Horizon None of these Pole Equator Horizon None of these ANSWER DOWNLOAD EXAMIANS APP
Surveying Closed contours, with higher value inwards, represent a None of these hillock plain surface depression None of these hillock plain surface depression ANSWER DOWNLOAD EXAMIANS APP
Surveying The displacement of the pictured position of a point of h elevation on a vertical photograph taken with a camera of 30 cm focal length, from an altitude of 3000 m, is 7.5 mm 6.5 mm 5.5 mm 4.4 mm 7.5 mm 6.5 mm 5.5 mm 4.4 mm ANSWER DOWNLOAD EXAMIANS APP
Surveying For adjusting a quadrilateral whose both the diagonals are observed, the equations of conditions involved, are One angle equation and three side equations None of these Three angle equations and one side equation Two angle equations and two side equations One angle equation and three side equations None of these Three angle equations and one side equation Two angle equations and two side equations ANSWER DOWNLOAD EXAMIANS APP
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